Exercise 2.1.30 (Last two digits of $3^{400}$)

Question

What are the last two digits in the ordinary decimal representation of $3^{400}$?

Richard Ganaye · Accepted Answer

\begin{proof} As in Exercise 29, we search the order of $3$ modulo $100$. By Euler's theorem,
$$3^{\phi(100)} \equiv 1 \pmod {100},$$
where $\phi(100) = \phi(2^2 5^2) = 100 \left(1 -\frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 40$. Thus
$$3^{40} \equiv 1 \pmod {100}.$$
($3^{40} = 12157665459056928801$.)
Then 
$$3^{400} = (3^{40})^{10} \equiv 1 \pmod {100}.$$
The last two digits of $3^{400}$ are $01$.
\end{proof}

Exercise 2.1.30 (Last two digits of $3^{400}$)

Answers

Comments

Exercise 2.1.30 (Last two digits of $3^{400}$)

Answers

Comments

Add answer